1.
Simple Dilution (Dilution Factor Method based on ratios)

A simple dilution is one in which
a unit volume of a liquid material of interest is combined
with an appropriate volume of a solvent liquid to achieve
the desired concentration. The dilution factor is the
total number of unit volumes in which your material will be dissolved.
The diluted material must then be thoroughly mixed to achieve
the true dilution. For example, a 1:5 dilution (verbalize as
"1 to 5" dilution) entails combining 1 unit volume
of solute (the material to be diluted) + 4 unit
volumes of the solvent medium (hence, 1 + 4 = 5
= dilution factor). The dilution factor is frequently expressed
using exponents: 1:5 would be 5e-1; 1:100 would be 10e-2, and
so on.

Example 1: Frozen orange juice concentrate is usually diluted
with 4 additional cans of cold water (the dilution solvent) giving
a dilution factor of 5, i.e., the orange concentrate represents
one unit volume to which you have added 4 more cans (same unit
volumes) of water. So the orange concentrate is now distributed
through 5 unit volumes. This would be called a 1:5 dilution,
and the OJ is now 1/5 as concentrated as it was originally. So,
in a simple dilution, add one less unit volume of solvent
than the desired dilution factor value.

Example 2: Suppose you must prepare 400 ml
of a disinfectant that requires 1:8 dilution from a concentrated
stock solution with water. Divide the volume needed by the dilution
factor (400 ml / 8 = 50 ml) to determine the unit volume. The
dilution is then done as 50 ml concentrated disinfectant + 350
ml water.

2.
Serial Dilution

A serial dilution is simply a series
of simple dilutions which amplifies the dilution factor quickly
beginning with a small initial quantity of material (i.e., bacterial
culture, a chemical, orange juice, etc.). The source of dilution
material (solute) for each step comes from the diluted material
of the previous dilution step. In a serial dilution the total
dilution factor at any point is the product of the
individual dilution factors in each step leading up to it.

Final dilution factor
(DF) = DF1 * DF2 * DF3
etc.

Example: In a typical microbiology exercise the students
perform a three step 1:100 serial dilution of a bacterial
culture (see figure below) in the process of quantifying the
number of viable bacteria in a culture (see figure below). Each
step in this example uses a 1 ml total volume. The initial step
combines 1 unit volume of bacterial culture (10 ul) with 99 unit
volumes of broth (990 ul) = 1:100 dilution. In the second step,
one unit volume of the 1:100 dilution is combined with
99 unit volumes of broth now yielding a total dilution of 1:100x100
= 1:10,000 dilution. Repeated again (the third step) the total
dilution would be 1:100x10,000 = 1:1,000,000 total dilution.
The concentration of bacteria is now one million times less
than in the original sample.

3.
Making fixed volumes of specific concentrations from liquid reagents:

V1C1=V2C2 Method

Very often you will need to make a specific
volume of known concentration from stock solutions, or perhaps
due to limited availability of liquid materials (some chemicals
are very expensive and are only sold and used in small quantities,
e.g., micrograms), or to limit the amount of chemical waste.
The formula below is a quick approach to calculating such dilutions
where:

V = volume, C = concentration; in whatever units
you are working.

(stock solution attributes) V1C1=V2C2 (new
solution attributes)

Example:
Suppose you have 3 ml of a stock solution of 100 mg/ml ampicillin
(= C1) and you want to make 200 ul (= V2) of
solution having 25 mg/ ml (= C2). You need to
know what volume (V1)
of the stock to use as part of the 200 ul total volume needed.

V1 = the volume of stock you will start with. This
is your unknown.
C1 = 100 mg/ ml in the stock solutionV2 = total volume needed at the new concentration
= 200 ul = 0.2 mlC2 = the new concentration = 25 mg/ ml

By algebraic rearrangement:

V1 = (V2 x C2)
/ C1

V1 = (0.2
ml x 25 mg/ml) / 100 mg/ml

and after cancelling the
units,

V1 = 0.05 ml, or 50 ul

So, you would take 0.05 ml = 50 ul
of stock solution and dilute it with 150 ul of solvent to
get the 200 ul of
25 mg/ ml solution needed. Remember that the amount of solvent
used is based upon the final volume needed, so you have to subtract
the starting volume form the final to calculate it.

Sometimes it may be more efficient to use molarity when expressing chemical concentrations. A mole is defined as exactly 6.023 x 1023 atoms, or molecules, of a substance (this is called Avagadro'snumber, N). The mass of one mole of an element is its atomic mass (g) and is noted for each element in the periodic table. Molecular weight is the mass (g) of a substance based on the summed atomic masses of the elements in the chemical formula. Formula weight refers to chemicals for which no discrete molecules exist; for example, NaCl in solid form is made up of Na+ and Cl- ions, but there are no true molecules of NaCl. The formulaweight of 1 mole NaCl would therefore be the sum of 1 atomic mass of each ion. The molecular weight (or FW) is provided as part of the information on the label of a chemical bottle. The number of moles in an arbitrary mass of an element or compound can be calculated as:

number of moles = weight (g)/ atomic (or molecular) weight (g)

Molarity (M) is the unit used to describe the number of moles of an element or compound in one liter (L) of solution (M = moles/L) and is thus a unit of concentration. By this definition, a 1.0 M solution is equivalent to one molecular weight (g/mole) of a compound brought up to 1 liter (1.0 L) volume with solvent (e.g., water) at a fixed temperature (liquids expand and contract with temperature and thus can change molarity).

Example1: To prepare a liter of a molar solution from a dry reagent

Multiply the molecular weight (or FW) by the desired molarity to determine how many grams of reagent to use:

Suppose a compound’s MW = 194.3 g/mole;

to make 0.15 M solution use 194.3 g/mole * 0.15 moles/L = 29.145g/L

You would dissolve the specified mass of reagent in a fraction of the total volume of solvent (at STP) and then raise the volume to exactly one liter by adding additional solvent and mixing thoroughly.

Example2: To prepare a specific volume of a specific molar solution from a dry reagent

A chemical has a FW of 180 g/mole and you need 25 ml (0.025 L) of 0.15 M (M = moles/L) solution. How many grams of the chemical are needed to make this solution?

Complex solutions such as buffers, salines, fixatives, etc., may be comprised of multiple chemical reagents. In preparation of these solutions, each reagent is dealt with separately in determining how much to use to make the final solution. For each, the volume used in the calculations is the final volume of solution needed.

Many reagents are mixed as percent solutions eitheras weight per volume (w/v) when starting with dry reagents OR volume per volume (v/v) when starting with liquid reagents. When preparing solutions from dry reagents, the same mass of any reagent is used to make a given percent concentration although the molar concentrations would be different.

Stock solutions of stable
compounds are routinely maintained in labs as more concentrated
solutions that can be diluted to working strength when used in
typical applications. The usual working concentration is denoted
as 1x. A solution 20 times more concentrated would be denoted
as 20x and would require a 1:20 dilution to restore the typical
working concentration.

Example: A 1x solution of a compound has a molar concentration
of 0.05 M for its typical use in a lab procedure. A 20x stock
would be prepared at a concentration of 20*0.05 M = 1.0 M. A
30X stock would be 30*0.05 M = 1.5 M.

7. Normality
(N): Conversion to Molarity

Normality = n*M where
n = number of protons (H+) in a molecule of the acid.